Question

Wayne wants to use Nitrogen, Potassium, and Phosphorus in his field as fertilizers. When any of them is mixed in the field, its quantity reduces by 1 kg every day due to chemical reactions. He mixed Nitrogen, Potassium, and Phosphorus on 7th November, 9th November, and 15th November, respectively. He spent equal amounts on buying each of the three. What should be the ratio of prices of Nitrogen, Potassium, and Phosphorus, so that there is an equal quantity of each of them in the field on 16th November, and that quantity is 11 kg?

Answer 1

Answer:

9:10:15

Step-by-step explanation:

On 16th November Nitrogen, Potassium, Phosphorus have the remaining quantity of 11 Kg each.  

Since every day the quantity of each element reduces by 1 Kg and Nitrogen added on 7th November, Potassium added on 9th November and Phosphorus added on 15th November.

So, the quantity of Nitrogen added on 7th November will be

11Kg+(16th-7th)*1Kg=11 Kg +9 Kg=20 Kg.

Again, the quantity of Potassium added on 9th November will be

11Kg+(16th-9th)*1Kg=11 Kg+ 7 Kg=18 Kg.

And, the quantity of Phosphorus added on 15th November will be

11Kg+(16th-15th)*1Kg= 11 Kg+1 Kg=12 Kg.

Given that Wayne spent an equal price for buying each 3 elements.

Let, x Rs. each is spent by him for buying those elements.

So, 20 Kg of Nitrogen cost x Rs.

⇒1 Kg of Nitrogen cost $\frac{x}{20}$ Rs.

Similarly, 18 Kg of Potassium cost x Rs.

⇒1 Kg of Potassium cost $\frac{x}{18}$ Rs.

Again, 12 Kg of Phosphorus cost x Rs.

⇒1 Kg of Phosphorus cost $\frac{x}{12}$ Rs.

Therefore, the required ratio of prices of Nitrogen, Potassium, and Phosphorus are $\frac{x}{20}:\frac{x}{18}:\frac{x}{12}$

=9:10:15 (Answer)